Green Yoda
Hi Φ
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- 2017
Find general solution to sin(3x)+sin(x)=0
Use the fact that sin(3x)=3sin(x)-4sin3(x) to complete this question.Find general solution to sin(3x)+sin(x)=0
An alternate method to leehuan's is to derive and use a sums-to-product identity.Find general solution to sin(3x)+sin(x)=0
wotUse the fact that sin(3x)=3sin(x)-4sin3(x) to complete this question.
Proof is left as an exercise to the reader.....
To derive that triple angle formula, you can apply the compound angle formula to sin(2x + x).
I tried that method..it didnt work out for meTo derive that triple angle formula, you can apply the compound angle formula to sin(2x + x).
sin(3x) = sin(2x + x)I tried that method..it didnt work out for me
They combine to give n.piAnswer only says nπ
but I also got nπ+(-1)^n x (π/2) or nπ+(-1)^n x (-π/2) as the gen sol
ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?They combine to give n.pi
Think about it.
(If you don't see it, start writing out the values for n=0,1,2,3,4,5,6,...)
Another approach would be to note that this equation is equivalent toFind general solution to sin(3x)+sin(x)=0
Doubt it.ah I see, just curious..in exam conditions if you dont pick up this connection..do you loose marks for writing 3 separate?
More motion question for you all (this will be good for the current 3U kids):
This is motion defined by a=a(x)
1. The acceleration of particle P, moving in a straight line, is given by x (with 2 dots on top of x)=2x-3 where x metres is the displacement from the origin O. Initially the particle is at O and its velocity v is 2 metres per second
a. Show that the velocity of the particle v^2=2x^2-6x+4
b. Calculate the velocity and acceleration of P at x=1 and briefly describe the motion of P after it moves from x=1
2. The acceleration of a particle is (2x-5) m/s^2, where x in the distance in metres from the origin
a. Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially
b. Describe the motion
3. A particle is moving along a straight line, with velocity v m/s and acceleration given by the expression k(4-x^2) m/s^2 where k is a constant
a. Show that v^2=4+Ae^(-2kx) where A is a constant satisifies the acceleration condition
b. If it starts from x=0 with a velocity of 7m/s, find the value of A
c. Does the particle ever change direction? Justify your answer
d. At x=1, v=4m/s, find the speed correct to two decimal places when x=2
e. As the motion continues, what happens, to the velocity and acceleration
cos(θ+10) = -1/2 [10<=θ+10<=370]I found it in my Maths in Focus text book. Also, I'll probably be posting innumerable seemingly easy questions considering I have an exam tomorrow haha. I've got another question: